The Frog Riddle - Conditional Probabilities

I saw this riddle doing the rounds on the internet: https://ed.ted.com/lessons/can-you-solve-the-frog-riddle-derek-abbott

In summary; There is a population of frogs with male:female occurring in 50:50 ratio. There are two patches of ground near you, one containing a single frog, the other containing two frogs. Your survival depends on you finding a female frog in one of these two patches, but you only get to make one attempt. You cannot tell which frogs are which in advance, except that you know that one of the frogs in the patch with two frogs in is male.

The answer given to the riddle is that the odds of the single frog being female is 50%, but the odds of one of the two frogs being female is 2/3 (67%). The explanation being that there are four possible combinations of male female pairs, one is excluded because we know one frog is male, hence 2/3 combinations where we find a female frog in the pair and 1/3 where we don't.

The probabilities just seem wrong to me; can anyone clarify the reason why this is the case?

I suspect that there is a subtly in the framing of the question that I'm missing.

As i read the problem, we have a choice of two options, both of which are simply a 50:50 chance of whether a single frog is male or female. Not knowing which frog in the pair is definitely male should have no effect on the probability of the other.

If I am wrong I really want to understand why!

• Can you restate the riddle here so readers don't have to follow the link (which also may break in the future) and then watch a video? Mar 13, 2016 at 23:11
• It seems to me that one has to make strong assumptions in order to obtain any answer. E.g., supposing male frogs croak only in the presence of a female, you would obtain one answer; but supposing that they tend to croak in the presence of another male, you would obtain a different answer (and make a different decision). Or what if females are not gregarious and tend to avoid other frogs? You would make yet a third decision. Although it's clearly intended that you ignore all such considerations, contemplating them may help you understand why the odds you compute are not necessarily 50:50.
– whuber
Mar 15, 2016 at 22:43
• The TED-Ed frog riddle answer is wrong. There is a very detailed answer here: duckware.com/tedfrog Jan 26, 2017 at 16:54
• @whuber You would also have to assume history did not exist, and that spotting a single male frog in a plurality of frogs was not evidence that that lone male frog had successfully sought out a female frog. There's all kinds of connoted assumptions about inter-sex relations as well, but those are possibly less germane to the probabilistic portion of the riddle. Mar 2, 2021 at 17:27

Let's look at the pair of frogs. Male frogs are identified by croaking in the video.

As explained in the video, before we hear any croaking, there are 4 equally likely outcomes given 2 frogs:

• Frog 1 is Male, Frog 2 is Male
• Frog 1 is Female, Frog 2 is Male
• Frog 1 is Male, Frog 2 is Female
• Frog 1 is Female, Frog 2 is Female

Making the assumptions about males and females occurring equally and independently, our sample space is $\{(M,M),(F,M),(M,F),(F,F)\}$, and we have probability $1/4$ for each element.

Now, once we hear the croak coming from this pair, we know that at least one frog is male. Thus the event $(F,F)$ is impossible. We then have a new, reduced sample space induced by this condition: $\{(M,M),(F,M),(M,F)\}$. Each remaining possibility is still equally likely, and the probability of all the events added together must be $1$. So the probability of each of these three events in the new sample space must be $1/3$.

The only event that ends badly for us is $(M,M)$, so there is a $2/3$ chance of survival.

More formally, the definition of conditional probability says:

$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$ So if $A$ is the event that at least one female is present and $B$ is the event that at least one male is present, we have: \begin{align}P(\text{F given at least 1 M}) &= \frac{P(\text{F and at least 1 male})}{P(\text{at least 1 M})}\\ &= \frac{P(\text{1 M and 1 F})}{P(\text{1 M or 2 M})} \\ &= \frac{P[(M,F),(F,M)]}{P[(M,M),(F,M),(M,F)]} \\ &= \frac{1/2}{3/4} = 2/3 \end{align}

This is really the same procedure we reasoned through as above.

• Hi mb7744, thanks for the quick response. I understand the answer as laid out, however this does look to me like double counting which is why I'm struggling to accept the answer. (M,F)=(F,M), surely, and if not, why? Mar 13, 2016 at 23:50
• (M,F) and (F,M) are not the same event. If one frog is named Alex and the other frog is named Taylor, Alex could be the female and Taylor the male OR vice versa. Alex and Taylor would probably disagree that this distinction is meaningless. Now, you could view the two events as equivalent. However, then your three outcomes (M,M),(F,F) and (M,F) are not equally likely. The mixed pairing is twice as likely. This is the same reason that you are much more likely to roll a 7 on a pair of dice than a 2, even if you view all of the different ways of rolling 7 as equivalent. Mar 13, 2016 at 23:57
• Hi, I think this helps clarify where i'm not 'getting' the riddle. If I may restate the problem as I'm seeing it, replace frog with a coin toss (or a dice roll). If you got to flip two coins and exclude certain combinations I would completely accept the answer. In the riddle's analogy however, i read this as we only get one coin toss. The other has already been made and cannot change the outcome of the other. Not knowing which of the two outcome has already been determined doesn't allow us to flip two coins and choose which outcomes to include or exclude. So using the dice roll analogy..... Mar 14, 2016 at 6:57
• ...you get to roll two dice, but unknown to you one dice' outcome has already been decided. You only have 1/6 chance of making any number 7-12. Am I wrong here? Mar 14, 2016 at 6:57
• If we look at all the pairs of equally-likely outcomes in dice rolling, order matters. Imagine one die is blue and the other red, and we write our outcomes with the blue die first and the red die last. Then the outcome (1,2) is not the same as the outcome (2,1). And, as before, the probability of rolling a "1 and a 2, regardless of order" will be twice as much as, say, rolling a pair of 2s. For your last question, I am assuming you meant to say one die's outcome was decided to be 6. In that case you are correct. Mar 14, 2016 at 12:50

Since the math is already laid out I'll try to provide some intuition. The issue is that knowing that at least one frog is male is different from knowing that any particular frog is male. The former case carries less information and this effectively increases our chances over the latter situation.

Call the frogs left and right, and suppose we are told that the right frog is male. Then we have eliminated two possible events from the sample space: the event where both frogs are female and the event where the left frog is male and the right frog is female. Now the probability truly is one half and it doesn't matter which one we choose. The exact same argument is true if we learn that the left frog is male.

But if we are told only that at least one frog is male, which is what happens when we hear the croak, then we cannot eliminate the event that the left frog is male and the right frog is female. We can only eliminate the event that both are female, which makes the event that at least one is female more likely than the previous setting.

I think the reason why this is confusing is that we naturally think learning that at least one is male should make us disinclined to choose the pair of frogs. It is true that this information makes it less probable that at least one is female, but recognize also that there was a full three quarters chance of at least one female before we learned anything at all. It's the ambiguity of the information we receive which makes it so we should still prefer the two frogs over the one.

• Thanks dsaxton, intuitively I opted for the two frogs, but my reasoning told me either choice was equally probable. Mar 14, 2016 at 7:02
• Thanks dsaxton, I suspect it's the phrasing of the riddle that is throwing me. As encountered, the two frogs are not distinguishable (without further information), so I am not seeing the (M,F), (F,M) distinction as meaningful in this context. I am not convinced that my reasoning is faulty, but my apologies if I am just being a bit slow. Mar 14, 2016 at 7:11
• Thanks again dsaxton. As mentioned above, I've found the mental hang up I was having and can see now why the answer is the right answer (and the question I was actually trying to answer). Thanks again for your help, seeing the answer is just not the same as having the help to really understand it. Mar 14, 2016 at 20:10

Your intuition is correct in this case. As the problem is stated your odds of survival are 50%. The video incorrectly states the problem space based on the information we have and therefore comes to an incorrect conclusion. The correct problem space contains 8 conditions and is as follows.

We have two frogs on a log, and one of them has croaked what are our possibilities? (M designates male, F designates female and c designates croaked, first position is left, second position is right)

[
[Mc, M],
[M, Mc],
[Mc, F],
[M, Fc], (X No Male croak)
[Fc, M], (X No Male croak)
[F, Mc],
[Fc, F], (X No Male croak)
[F, Fc], (X No Male croak)
]


Each case is equally likely based on the information that we have, when we eliminate the conditions given the knowledge that a male frog has croaked. We find that there are 4 outcomes to expect. Left male frog croaked next to a right male frog that was silent. Right male frog croaked next to a left male frog that was silent. Or there was a croaking male frog paired with a single female frog in either direction. For an intuitive way to understand this, the two male frogs are twice as likely to croak than the single male frog paired with a female, so we have to weight it appropriately.

You could also divide the search space by croaking frog (C) and non croaking frog (N). Since the croaking frog is 100% a male, you can eliminate it from your search since it has no chance of helping you survive. While the author intended to create a "monty hall problem" they inadvertently created a "boy or girl paradox".

The following questions yield different results:

Given that there is a male what is the likelihood the other is female?

Given that a male frog croaked what is the likelihood the other is female?

https://en.wikipedia.org/wiki/Monty_Hall_problem

A clearer answer to this, since the previous was too long and not easy to understand.

The possible outcomes are different, although I used same letters. To make clear the sample space, I will describe the possible outcomes

M M --> "The male is on the left" - "A random male on the right"

M F --> "The male is on the left" - "A random female on the right"

M M --> "The male is on the right" - "A random male on the left"

M F --> "The male is on the right" - "A random female on the left"

• You are double counting the MM case. You can't just enumerate all the possible scenarios without taking into account whether you're arriving at the same scenario through different paths. Jan 8, 2019 at 23:20

The problem I have with this problem, is that the solution seems to be using different rules for what it considers a possible result for the two frogs being male and female, and male and male.

The F/M pair, and the M/F pair, are different because we don't know whether the first frog or the second frog is male, so F/M and M/F are two separate possibilities, even though the result still amounts to "one female frog, one male frog".

But the M/M pair is only considered one possible result, even though the same logic should apply: we don't know which frog is the one that made the croaking sound, so either frog could be the one we heard, and the other one could still be male, it just didn't happen to croak.

• This is more in the nature of a comment than an answer to the "riddle." Please change it to a comment and delete this "answer." Apr 1, 2016 at 11:32
• @DJohnson Actually, this is an answer to the riddle, although the later answer from tomciopp explains it more clearly. Jan 8, 2019 at 23:18

Not knowing anything: $$\{(M,M), (M,F), (F,M), (F,F)\}$$. Three pairs with at least one female out of four possible combinations: $$3/4$$ or $$75\%$$

Knowing the first one is male: $$\{(M,M), (M,F)\}$$. One pair with at least one female out of two possible combinations: $$1/2$$ or $$50\%$$

Knowing that there is at least one male: $$\{(M,M), (M,F), (F,M)\}$$. Two pairs with at least one female out of three possible combinations: $$2/3$$ or $$67\%$$

The correct answer is given above by tomciopp and the video is incorrect. I want to elaborate on their answer with a little more to diagnose the video's mistake and give an intuitive illustration or two.

The Video's Mistake

In the video, they conditionalize on the statement "at least one frog is male." However, the information we receive is "there was exactly one croak in the clearing." This statement implies that there is at least one male frog in the clearing, but it is not logically equivalent to it. If exactly one croaked, then at least one is male; however, "if at least one is male, exactly one croaked" is false.

If we conditionalize on the information we receive (instead of a statement which it merely implies) tomciopp's analysis is correct.

An intuitive illustration

Here's an intuitive case to help understand the difference. Suppose I flip two coins and exactly one of them is a penny. Then I ask you "what's the probability that the non-penny is a heads?" The answer is 0.5. It doesn't matter what the penny did. If I add a piece of information like "All the pennies I have ever flipped were heads" that doesn't change one whit your expectation about non-pennies.

In the case of the clearing, we have a croaker (=penny) and a non-croaker (=non-penny.) We also know that croakers are irrelevant to our decisions. 50% of non-croakers are female. The case is symmetric to coins. (This is sort of implied in the problem. Technically, if a frog hasn't croaked for a certain amount of time, I probably start to think it is more likely to be female, since males sometimes croak. In this case, however, we can fairly ignore this since each non-croaking frog has been observed the same about of time, they have the same increasing probability of being female.)

A Different Case

There is a species of toad that prevents mushroom poisoning. Only the females of this species have the necessary secretion to prevent mushroom poisoning. The females of this species do a funny dance whenever there is another female around, but not otherwise.

In this case, the pair is not dancing if and only if at least one is male. Thus, we can conditionalize on "at least one is male" and validly draw the conclusion there is a 2/3 probability of a female in a non-dancing pair.

Before we hear any croaking, there are 4 equally likely outcomes given 2 frogs:

Frog 1 is Male, Frog 2 is Male

Frog 1 is Female, Frog 2 is Male

Frog 1 is Male, Frog 2 is Female

Frog 1 is Female, Frog 2 is Female

Making the assumptions about males and females occurring equally and independently, our sample space is {(M,M),(F,M),(M,F),(F,F)}, and we have probability 1/4 for each element.

Once we hear the croak coming from this pair, we know that at least one frog is male. This male can equally likely be Frog 1 or Frog 2. So there are 2 equally likely outcomes for the Frog 1:

Frog 1 is Male

Frog 1 is Random Frog

Making the assumptions about males and females occurring equally and independently, the Random Frog is equally likely to be a Random Male or a Random Female.

P(Frog 1 is Random Male given Frog 1 is Random Frog)=P(Frog 1 is Random Female given Frog 1 is Random Frog)=1/2

P(Frog 1 is Random Male and Frog 1 is Random Frog)=P(Frog 1 is Random Frog)P(Frog 1 is Random Male given Frog 1 is Random Frog)=(1/2)(1/2)=1/4

P(Frog 1 is Random Female and Frog 1 is Random Frog)=P(Frog 1 is Random Frog)P(Frog 1 is Random Female given Frog 1 is Random Frog)=(1/2)(1/2)=1/4

So there are 3 possible outcomes for the Frog 1:

Frog 1 is Male

Frog 1 is Random Male

Frog 1 is Random Female

and probabilities are:

P(Frog 1 is Male)=1/2

P(Frog 1 is Random Male)=1/4

P(Frog 1 is Random Female)=1/4

Now, for each possible outcome for Frog 1, there are 2 possible outcomes for the Frog 2:

Frog 2 is Male

Frog 2 is Random Frog

For each possible outcome for Frog 1, the Random Frog is equally likely to be a Random Male or a Random Female.

So, for each possible outcome for Frog 1, there are 3 possible outcomes for the Frog 2:

Frog 2 is Male

Frog 2 is Random Male

Frog 2 is Random Female

P(Frog 2 is Male given Frog 1 is Male)=0

P(Frog 2 is Male given Frog 1 is Random Male)=1

P(Frog 2 is Male given Frog 1 is Random Female)=1

P(Frog 2 is Random Male given Frog 1 is Male)=1/2

P(Frog 2 is Random Male given Frog 1 is Random Male)=0

P(Frog 2 is Random Male given Frog 1 is Random Female)=0

P(Frog 2 is Random Female given Frog 1 is Male)=1/2

P(Frog 2 is Random Female given Frog 1 is Random Male)=0

P(Frog 2 is Random Female given Frog 1 is Random Female)=0

P(Frog 2 is Random Male and Frog 1 is Male)=P(Frog 1 is Male)P(Frog 2 is Random Male given Frog 1 is Male)=(1/2)(1/2)=1/4

P(Frog 2 is Random Female and Frog 1 is Male)=P(Frog 1 is Male)P(Frog 2 is Random Female given Frog 1 is Male)=(1/2)(1/2)=1/4

P(Frog 2 is Male and Frog 1 is Random Male)=P(Frog 1 is Random Male)*P(Frog 2 is Male given Frog 1 is Random Male)=(1/4)*1=1/4

P(Frog 2 is Male and Frog 1 is Random Female)=P(Frog 1 is Random Female)*P(Frog 2 is Male given Frog 1 is Random Female)=(1/4)*1=1/4

So, our sample space is {(Male,Random Male),(Male,Random Female),(Random Male,Male),(Random Female,Male)}, and we have probability 1/4 for each element.

P(F given at least 1 M)=P(F and at least 1 male)/P(at least 1 M)=P(1 M and 1 F)/P(1 M or 2 M)=P[(Male,Random Female),(Random Female,Male)]/P[(Male,Random Male),(Male,Random Female),(Random Male,Male),(Random Female,Male)]=(1/2)/(4/4)=1/2

• Did you copy and paste from my answer and remove the formatting? Apr 1, 2016 at 13:01
• Well, first of all, copying and pasting a part of someone else's answer without even mentioning it is unacceptable. That aside, if you think that you have reached a different result, is there a more concise way for you to explain it? You have written a lot of disconnected equations without any explanation. Apr 2, 2016 at 4:16
• It's not literature but it is still rude. Now, with regards to your answer versus mine: I find yours nonsensical. What is the meaning of the outcome "Frog 2 is Random Frog"? Apr 2, 2016 at 17:31
• Your answer was the only one calculating conditional probabilities. Using same terms could help comparing and see which part is the same and which is different. I may say, I find other answers nonsensical too, but I didnt say so because it would be rude ;) . If you dont understand sth, you can jusk ask for clarifications. "Frog 2 is Random Frog" means it is not the male frog known to be in the pair.... Apr 2, 2016 at 18:52
• There are two sources of randomness, one coming from the male frog known to be in the pair, the other coming from the frog population. Since we know the male frog is there, the uncertainty is just about the position. Is it frog 1 or frog 2? Or, is it on the left or on the right? My advice is, use tree diagram to build sample space from scratch and use all info available. Apr 2, 2016 at 18:52